1.- When the gaseous reaction N204 ‹> 2NO2 is carried out at 328 K, a Kc value of 0.05 corresponds to it. If 92 g of NO2 are initially fed to a reactor with a capacity of 18 liters so that the corresponding reaction is carried out at the same temperature of 328 K, calculate: a.- The yield of the reaction when reaching equilibrium. (in %) b. - The total pressure at equilibrium (in atm) c.- The composition in % mol of the mixture in equilibriumm% n_(N204)

A batch process is operated with the following sequence. The feed to the batch process consists of a solution of 500 L of ethanol with 120 mole of A and 120 mole of B. If the temperature of the solution is sufficient, A and B will react to form P (the desired product): A (eth) + B (eth) → P (s) DHreact  = − 250 kJ/mol (1)  product P is sparingly soluble in ethanol, which helps facilitate its removal from the reaction solution. It can be assumed that the reaction rate is first-order in both the concentration of A (cA ) and B (cB ), and so second-order overall: -r A = k c A cB  (2) The batch process consists of the following FOUR steps: Initially, the tank is empty. 1. Fill tank with solution (ethanol/A/B). 50°C (1 hour) Increase temperature of the feed from 20°C to 2. Reaction takes place isothermally at 50°C (4 hours, to give conversion of 50%) 3. Filtration of reaction solution to give product P (2 hours) 4. Drying with hot nitrogen gas followed by the removal of powder product (3…

The formation of NO from Na and O is to be carried out in a small batch reactor. As a first approximation, we shall consider that the reaction time is more rapid than the time of cylinder compression, consequently, the reaction takes place isothermally in car eylinder at 2700 K, in a constant volume reactor (cylinder) and under a pressure of 20 atm, the initial concentration of N2 is 0.0696 mol/liter. By a specifying constant volume, we are assuming that reaction take place rapidly with respect to the movement of the piston in the cylinder. Consider that the feed consists of 77% N2, 15% Oz and 8% other gases, which may be considered inert. At this temperature the equilibrium constant (Ke-0.01). The reaction is reversible: Na + 0z + 2NO With a rate equation: Cho -TN, = k ( CN, Co, Ke Calculate: a) The equilibrium conversion of N2. b) The time required to achieve 80% of the equilibrium conversion. The formation reaction rate- constant k at this temperature is 1.11 liter mol.h' Hint: dx…

Ethane (C2H6) gas at mass flow rate of 1.5 kg/min is mixed with air (O2 + 3.76N2) at mass flow rate of 1 kg/sec in a steady-state, steady-flow isobaric combustion process. All reactants enter the process at 298 K and 100 kPa, and the fuel burns to complete combustion. a) Determine the Air-Fuel Ratio (AFR) b) Write a balanced reaction equation based on one mole of fuel c) Determine the excess air fraction (%) d) Assuming that the combustion products exit the system at 1000 K, determine the volumetric flow rate of the combustion exhaust (m3 /s) e) Assuming that the combustion products exit the system at 1000 K, determine the rate of heat transfer to/from the system (kW)Repeat steps d and e, but assuming that the combustion product exit the system at 2000 k. Using the values in part e for 1000 k and 2000 k, apply linear interpolation to estimate the adiabatic exit temperature (?̇ = 0) for the combustion products. Next, using conservation of energy analysis, solve for the adiabatic exit…